# A limits question

The following equals? $$\lim_{x \to 1}\frac{\displaystyle\int_1^x \sin(t) \, dt}{x^2-1}$$ I think this can be converted to $$\lim_{x \to 1}\frac{\sin(x)}{2x} = \frac{\sin(1)}{2}$$ by using the fundamental theorem of calculus.
But the correct answer is $1/2$.
So where I made a mistake?

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Do you mean this? $$\lim_{x \to 1} \frac{\int_1^x \sin t dt}{x^2 - 1}$$ –  user61527 Apr 12 '14 at 22:15
Yes..I don't know how to type it that way sorry –  shidangai Apr 12 '14 at 22:15
You can definitely use l'Hôpital's theorem and the fundamental theorem of calculus; your computation seems good. –  egreg Apr 12 '14 at 22:17

$$\lim_{x \to 1} \frac{1}{x + 1} \cdot \frac{\int_1^x \sin t}{x - 1}$$
$$\frac 1 {1 + 1} \cdot \frac{d}{dx} \left(\int_1^x \sin t dt\right)\Big|_{x = 1} = \frac 1 2 \sin 1$$